∫(x^2)/(≤ft(x^2+1)≤ft(x^2+9)) dx is equal to

The correct answer is Option (1) → \(\displaystyle -\frac{1}{8}\tan^{-1}x+\frac{3}{8}\tan^{-1}\left(\frac{x}{3}\right)+C\)

Only \(x^2\) appears, so put \(y = x^2\) for the partial fractions step:

    \( \displaystyle \frac{y}{(y+1)(y+9)} = \frac{A}{y+1}+\frac{B}{y+9} \)

    \( y = A(y+9)+B(y+1) \)

Putting y = −1 and y = −9:

    \( \displaystyle A = -\frac{1}{8},\quad B = \frac{9}{8} \)

So

    \( \displaystyle \int\frac{x^2\,dx}{\left(x^2+1\right)\left(x^2+9\right)} = -\frac{1}{8}\int\frac{dx}{x^2+1}+\frac{9}{8}\int\frac{dx}{x^2+9} \)

Using \(\int\frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}\):

    \( \displaystyle = -\frac{1}{8}\tan^{-1}x+\frac{9}{8}\times\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)+C \)

    \( \displaystyle = -\frac{1}{8}\tan^{-1}x+\frac{3}{8}\tan^{-1}\left(\frac{x}{3}\right)+C \)

Why the other options are incorrect

• Option (2) \(\displaystyle -\frac{1}{8}\tan^{-1}x+\frac{9}{8}\tan^{-1}\left(\frac{x}{3}\right)+C\) — Incorrect. It forgets the factor \(\frac{1}{a} = \frac{1}{3}\) in \(\int\frac{dx}{x^2+9}\).

• Option (3) \(\displaystyle\frac{1}{8}\tan^{-1}x-\frac{3}{8}\tan^{-1}\left(\frac{x}{3}\right)+C\) — Incorrect. Both signs are reversed; this is the negative of the answer.

• Option (4) \(\displaystyle -\frac{1}{8}\tan^{-1}x+\frac{1}{8}\tan^{-1}\left(\frac{x}{3}\right)+C\) — Incorrect. B is \(\frac{9}{8}\), so the second coefficient is \(\frac{9}{8}\times\frac{1}{3} = \frac{3}{8}\), not \(\frac{1}{8}\).